Answer to Question #156842 in Calculus for joel

Question

A projectile is fired from the origin in O with initial velocity(5i + 29j)m/s. The projectile just clears a wall of height 20m which is at a horizontal distance x from O. The plane of the wall is perpendicular to the vector i. Find the possible values of x. Find also the corresponding values of the velocity of the projectile when it is just passing over the wall.

Answer 1

Solution

v_0x = 5, v_0y = 29

Velocity at any time t: v_x(t) = v_0x, v_y(t) = v_0y - gt (g - acceleration earth gravity)

Therefore x(t) = v_0x *t, y(t) = v_0y*t - g*t²/2

A projectile pass height h = 20 at the time which satisfy equation h = v_0y*t - g*t²/2

t_1,2 = (v_0y ±√( v_0y² – 2gh))/g

t₁ = (v_0y -√( v_0y² – 2gh))/g = 0.797 s, t₂ = (v_0y +√( v_0y² – 2gh))/g = 5.117 s

So x₁ = v_0x *t₁ = 3.985 m,  x₂ = v_0x *t₂ = 25.586 m

v_1x = v_0x = 5 m/s,   v_1y = v_0y – gt₁  = 21.183 m/s, v₁ = √( v_1x² + v_1y² ) = 21.765 m/s

v_2x = v_0x = 5 m/s,   v_2y = v_0y – gt₂  = -21.183 m/s, v₂ = √( v_2x² + v_2y² ) = 21.765 m/s

Answer

x₁ = 3.985 m, x₂ = 25.586 m

v_1x = 5 m/s,  v_1y = 21.183 m/s, v₁ = 21.765 m/s

v_2x = 5 m/s,  v_2y = -21.183 m/s, v₂ = 21.765 m/s